The ramjet cycle. Chapter Ramjet flow field

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1 Chapter 3 The ramjet cycle 3. Ramjet flow field Before we begin to analyze the ramjet cycle we will consider an example that can help us understand how the flow through a ramjet comes about. The key to understanding the flow field is the intelligent use of the relationship for mass flow conservation. In this connection there are two equations that we will rely upon. The first is the expression for -D mass flow in terms of the stagnation pressure and temperature. ṁ = UA = + + 2( ) 2 The second is the all-important area-mach number function. Pt A p f (M) (3.) RTt f (M) = A A = ( ) M M 2 2( ) (3.2) This function is plotted in Figure 3. for three values of. For adiabatic, isentropic flow of a calorically perfect gas along a channel Equation (3.) provides a direct connection between the local channel cross sectional area and Mach number. In addition to the mass flow relations there are two relationships from Rayleigh line theory that are also very helpful in guiding our understanding of the e ect of heat addition on 3-

2 CHAPTER 3. THE RAMJET CYCLE 3-2 Figure 3.: Area - Mach number relation. the flow in the ramjet. These are the equations that describe the e ect of heat addition on the Mach number and stagnation pressure of the flow. T t T t = P t P t = + M 2 2 2(+ ) M M 2 + M 2 + These equations are plotted in Figure 3.2. (3.3)! M 2 Figure 3.2: E ects of heat exchange on Mach number and stagnation pressure. There are several features shown in these plots that have important implications for the ramjet flow. The first is that much more heat can be added to a subsonic flow than to a supersonic flow before thermal choking occurs; that is, before the flow is brought to Mach one. The second is that stagnation pressure losses due to heat addition in subsonic flow are relatively small and cannot exceed about 20% of the stagnation pressure of the flow

3 CHAPTER 3. THE RAMJET CYCLE 3-3 entering the region of heat addition. In contrast stagnation pressure losses due to heat addition can be quite large in a supersonic flow. With this background we will now construct a ramjet flow field beginning with supersonic flow through a straight, infinitely thin tube. For definiteness let the free stream Mach number be three and the ambient temperature T 0 = 26 K. Throughout this example we will assume that the friction along the channel wall is negligible. Figure 3.3: Step - Initially uniform Mach three flow. Add an inlet convergence and divergence. Figure 3.4: Step 2 - Inlet convergence and divergence with f(m) shown. Let the throat Mach number be two (M.5 =2.0). In Figure 3.4 the Mach number decreases to the inlet throat (f(m) increases), then increases again to the inlet value of three, (f(m) = 0.236). The thrust of this system is clearly zero since the x-directed component of the pressure force on the inlet is exactly balanced on the upstream and downstream sides of the inlet. In Figure 3.5 heat is added to the supersonic flow inside the engine. Neglect the mass flow of fuel added compared to the air mass flow. As the heat is added the mass flow is conserved. Thus, neglecting the fuel added, the mass balance is

4 CHAPTER 3. THE RAMJET CYCLE 3-4 Figure 3.5: Step 2 - Introduce a burner and add heat to the flow. ṁ = + + 2( ) 2 P t0 A p 3 f (3) = RTt ( ) P t4 A 4 p RTt4 f (M 4 ). (3.4) As the heat is added, T t4 goes up and P t4 goes down while the following equality must be maintained P t0 p Tt0 f (3) = P t4 p Tt4 f (M 4 ). (3.5) Conservation of mass (3.5) implies that f(m 4 ) must increase and the Mach number downstream of the burner decreases. There is a limit to the amount of heat that can be added to this flow and the limit occurs when f(m 4 ) attains its maximum value of one. At this point the flow looks like Figure 3.6. The Rayleigh line relations tell us that the temperature rise across the burner that produces this flow is T t4 =.53. (3.6) T t3 M 4 = The corresponding stagnation pressure ratio across the system is

5 CHAPTER 3. THE RAMJET CYCLE 3-5 Figure 3.6: Step 3 - Introduce su slightly greater than one. cient heat to bring the exit Mach number to a value P t4 P t3 M 4 =! before unstart = (3.7) Now, suppose the temperature at station 4 is increased very slightly. We have a problem; T t4 is up slightly, P t4 is down slightly but f(m 4 ) cannot increase. To preserve the mass flow rate imposed at the inlet, the supersonic flow in the interior of the engine must undergo an un-start. The flow must switch to the configuration shown in Figure 3.7. The mass flow equation (3.5) can only be satisfied by a flow between the inlet throat and the burner that achieves the same stagnation pressure loss (3.7), since f(m 4 ) cannot exceed one and the stagnation temperature ratio is essentially the same. As a result of the un-start, a shock wave now sits at the end of the di user section of the inlet. Notice that the engine internal pressure is still very large and the exit Mach number must remain one. The stagnation temperature has not changed and so, as was just pointed out, the mass balance tells us that the stagnation pressure of the exit flow must be the same as before the un-start. Thus P t4 P t0 M 4 =! after unstart = (3.8) Now, the stagnation pressure loss is divided between two mechanisms, the loss across the

6 CHAPTER 3. THE RAMJET CYCLE 3-6 Figure 3.7: Step 4 - Increase the heat added very slightly to unstart the flow. shock wave, and the loss due to heat addition across the burner. The stagnation pressure ratio across a Mach three shock wave is P t3 = (3.9) P t0 M=3 The burner inlet Mach number is M 3 =0.475, and the stagnation loss due to thermal choking across the burner is The product of (3.9), and (3.0) is P t4 = (3.0) P t3 M=0.475 Now let s look at the thrust generated by the flow depicted in Figure 3.7. definition, neglecting the fuel/air ratio, is The thrust The pressure ratio across the engine is T 2 Ue = M 0 + A e Pe. (3.) P 0 A 0 U 0 A 0 P 0

7 CHAPTER 3. THE RAMJET CYCLE 3-7 and the temperature ratio is P e = P te + 2 M 0 2 P 0 P t0 + This produces the velocity ratio 2 M e 2 T e = T te + 2 M! 0 2 T 0 T t0 + 2 M e 2! =0.292 = =5.66 (3.2) = (3.3).2 Now substitute into (3.). U e = M r e Te = (3.4) U 0 M 0 T 0 T P 0 A 0 =.4 9 ( ) + (5.66 ) = = 0 (3.5) The thrust is zero. We would expect this from the symmetry of the upstream and downstream distribution of pressure on the inlet. Now let s see if we can produce some thrust. First adjust the inlet so that the throat area is reduced until the throat Mach number is just slightly larger than one. This will only e ect the flow in the inlet and all flow variables in the rest of the engine will remain the same. With the flow in the engine subsonic, and the shock positioned at the end of the di user, we have a great deal of margin for further heat addition. If we increase the heat addition across the burner the mass balance (3.5) is still preserved and the exit Mach number remains one. Let the burner outlet temperature be increased to T t4 = 84.4K. The flow now looks something like Figure 3.8. The stagnation temperature at the exit is up, the stagnation pressure is up, and the shock has moved to the left to a lower upstream Mach number (higher f(m) ), while the mass flow (3.5) is preserved. Note that we now have some thrust arising from the x-component of the high pressure force behind the shock that acts to the left on an outer portion of the inlet surface. This pressure exceeds the inlet pressure on the corresponding upstream portion of the inlet surface. The stagnation pressure ratio across the engine is determined from the mass balance (3.5). P te P t0 = f (3) r Tte T t0 =0.263 p 3=0.409 (3.6)

8 CHAPTER 3. THE RAMJET CYCLE 3-8 Figure 3.8: Step 5 - Increase the heat addition to produce some thrust. Let s check the thrust. The pressure ratio across the engine is The temperature ratio is and the velocity ratio is now P e =0.409 =7.94. (3.7) P 0.2 T e = T te + 2 M! 0 2 T 0 T t0 + 2 M e 2 =3 2.8 = 7 (3.8).2 The thrust is U e = M r p e Te 7 = = (3.9) U 0 M 0 T 0 3 T P 0 A 0 =.4 9 (0.882 ) + (7.94 ) = = (3.20) This is a pretty substantial amount of thrust. Note that, so far, the pressure term in the thrust definition is the important thrust component in this design.

9 CHAPTER 3. THE RAMJET CYCLE The role of the nozzle Let s see if we can improve the design. Add a convergent nozzle to the engine as shown in Figure 3.9. The mass balance is Figure 3.9: Step 6 - Add a convergent nozzle. P t0 A.5 p = P tea e p (3.2) How much can we decrease A e? Begin with Figure 3.8. As the exit area is decreased the exit Mach number remains one due to the high internal pressure in the engine. The shock moves upstream toward the inlet throat, the exit stagnation pressure increases and the product P te A e remains constant. The minimum exit area that can be reached without un-starting the inlet flow is when the inlet shock is very close to the throat, becoming vanishingly weak. At this condition, the only mechanism for stagnation pressure loss is the heat addition across the burner. The Mach number entering the burner is M 3 =0.38 as shown in Figure 3.9. The stagnation pressure loss across the burner is proportional to the square of the entering Mach number. dp t P t = M 2 dt t T t (3.22) To a reasonable approximation the stagnation loss across the burner can be neglected and we can take P te = Pt0. In this approximation, the area ratio that leads to the flow depicted

10 CHAPTER 3. THE RAMJET CYCLE 3-0 in Figure 3.9 is r A e 84.4 = =.732. (3.23) A.5 ideal This relatively large area ratio is expected considering the greatly increased temperature and lower density of the exhaust gases compared to the gas that passes through the upstream throat. What about the thrust? Now the static pressure ratio across the engine is P e P 0 = = 9.4. (3.24).2 The temperatures and Mach numbers at the nozzle exit are the same as before so the velocity ratio does not change between Figure 3.8 and Figure 3.9. The dimensionless thrust is T P 0 A 0 =.4 9 (0.882 ) (9.4 ) = = (3.25) That s pretty good; just by adding a convergent nozzle and reducing the shock strength we have increased the thrust by about 20%. Where does the thrust come from in this ramjet design? The figure below schematically shows the pressure distribution through the engine. The pressure forces on the inlet and nozzle surfaces marked a roughly balance, although the forward pressure is slightly larger compared to the rearward pressure on the nozzle due to the heat addition. But the pressure on the inlet surfaces marked b are not balanced by any force on the nozzle. These pressures substantially exceed the pressure on the upstream face of the inlet and so net thrust is produced. 3.3 The ideal ramjet cycle But we can do better still! The gas that exits the engine is at a very high pressure compared to the ambient and it should be possible to gain thrust from this by adding a divergent section to the nozzle as shown below. The area ratio of the nozzle is chosen so that the flow is fully expanded, P e = P 0. The stagnation pressure is constant through the engine and so we can conclude from

11 CHAPTER 3. THE RAMJET CYCLE 3- Figure 3.0: Imbalance of pressure forces leading to net thrust. Figure 3.: Ideal ramjet with a fully expanded nozzle.

12 CHAPTER 3. THE RAMJET CYCLE 3-2 that M e = M 0. The temperature ratio is Finally, the dimensionless thrust is P e = P te + 2 M 0 2 P 0 P t0 + 2 M e 2! (3.26) = + 2 M! M e 2 T e = T te + 2 M! 0 2 T 0 T t0 + 2 M e 2 = T te T t0 =3. (3.27) r! T 2 M e Te p = M 0 = P 0 A 0 M 0 T 0 =9.22. (3.28) Adding a divergent section to the nozzle at this relatively high Mach number increases the thrust by 50%. Now work out the other engine parameters. The fuel/air ratio is determined from ṁ f h f =(ṁ a +ṁ f ) h t4 ṁ a h t3. (3.29) Assume the fuel added is JP-4 with h f = J/kg. Equation (3.29) becomes f = T t4 T t3 h f C pt t3 = T t4 T t = (3.30) The relatively small value of fuel/air ratio is the a posteriori justification of our earlier neglect of the fuel mass flow compared to the air mass flow. If we include the fuel/air ratio in the thrust calculation (but still ignore the e ect of mass addition on the stagnation pressure change across the burner) the result is T P 0 A 0 = M 0 2 ( + f) M r! e Te = p 3 = (3.3) M 0 T 0

13 CHAPTER 3. THE RAMJET CYCLE 3-3 The error in the thrust is about 7% when the fuel contribution is neglected. The dimensionless specific impulse is I sp g a 0 = f M 0 T P 0 A 0 = and the overall e ciency is, ( f = h f /C p T 0 = 97.2), = 79.4 (3.32) ov = f f T P 0 A 0 = = (3.33) The propulsive e ciency is pr = 2U 0 U e + U 0 = 2 + p 3 = (3.34) The thermal e ciency of the engine shown in Figure 3. can be expressed as follows th = (ṁ a +ṁ f ) Ue 2 U 2 2 ṁ 0 a 2 = (ṁ a +ṁ f )(h te h e ) ṁ a (h t0 h 0 ) ṁ f h f (ṁ a +ṁ f ) h te ṁ a h t0 th = Q rejected during the cycle Q input during the cycle = (ṁ a +ṁ f ) h e ṁ a h 0 (ṁ a +ṁ f ) h te ṁ a h t0 (3.35) th = T 0 T t0! ( + f) Te T 0. ( + f) Tte T t0 The heat rejection is accomplished by mixing of the hot exhaust stream with surrounding air at constant pressure. Noting (3.27) for the ideal ramjet, the last term in brackets is one and the thermal e ciency becomes th ideal ramjet = T 0 T t0. (3.36) For the ramjet conditions of this example the thermal e e ciency is ciency is 2/3. The Brayton cycle B = T 0 T 3. (3.37)

14 CHAPTER 3. THE RAMJET CYCLE 3-4 In the ideal cycle approximation, the Mach number at station 3 is very small thus T 3 = Tt0 and the thermal and Brayton e ciencies are identical. Note that, characteristically for a Brayton process, the thermal e ciency is determined entirely by the inlet compression process. The ramjet design shown in Figure 3. represents the best we can do at this Mach number. In fact the final design is what we would call the ideal ramjet. The ideal cycle will be the basis for comparison with other engine cycles but it is not a practically useful design. The problem is that the inlet is extremely sensitive to small disturbances in the engine. A slight increase in burner exit temperature or decrease in nozzle exit area or a slight decrease in the flight Mach number will cause the inlet to un-start. This would produce a strong normal shock in front of the engine, a large decrease in air mass flow through the engine and a consequent decrease in thrust. A practical ramjet design for supersonic flight requires the presence of a finite amplitude inlet shock for stable operation. 3.4 Optimization of the ideal ramjet cycle For a fully expanded nozzle the thrust equation reduces to T P 0 A 0 = M 0 2 ( + f) M r! e Te. (3.38) M 0 T 0 For the ideal cycle, where P te = P t0, M e = M 0 and T te /T t0 = T e /T 0, the thrust equation using ( + f) =( f r ) / ( f ) becomes T = 2 P 0 A 0 ( f r ) f r r. (3.39) r This form of the thrust equation is useful because it expresses the thrust in terms of cycle parameters that we can rationalize. The parameter r is fixed by the flight Mach number. At a given altitude is determined by maximum temperature constraints on the hot section materials of the engine, as well as fuel chemistry, and gas dissociation. If the flight Mach number goes to zero the thrust also goes to zero. As the flight Mach number increases for fixed the fuel flow must decrease until = r when fuel shut-o occurs and the thrust is again zero. A typical thrust plot is shown below. The optimization question is; at what Mach number should the ramjet operate for maximum thrust at a fixed? Di erentiate (3.39) with respect to r and set the result to zero.

15 CHAPTER 3. THE RAMJET CYCLE 3-5 Figure 3.2: Ramjet thrust 0 T = r 3 r +2 r + r f + r r P 0 A 0 ( f ) 2 r r The value of r for maximum thrust is determined from 2 r 2 q r A =0 (3.40) r r 2 r 3 r +2 r + f + r 2 r =0. (3.4) r r The quantity f is quite large and so the second term in parentheses in (3.4) clearly dominates the first term. For f the maximum thrust Mach number of a ramjet is found from /2 = 2( r max thrust ) 3/2 rmax thrust +. (3.42) For the case shown above, with =8.4, the optimum value of r is 3.5 corresponding to a Mach number of The ramjet is clearly best suited for high Mach number flight and the optimum Mach number increases as the maximum engine temperature increases. I sp g a 0 = 2 ( r ) /2 f r f f r r r (3.43)

16 CHAPTER 3. THE RAMJET CYCLE 3-6 The specific impulse also has an optimum but it is much more gentle than the thrust optimum, as shown in Figure 3.3. Figure 3.3: Ramjet specific impulse. Optimizing the cycle with respect to thrust essentially gives close to optimal specific impulse. Notice that the specific impulse of the ideal cycle has a finite limit as the fuel flow reaches shut-o. 3.5 The non-ideal ramjet The major non-ideal e ects come from the stagnation pressure losses due to the inlet shock and the burner heat addition. We have already studied those e ects fairly thoroughly. In addition there are stagnation pressure losses due to burner drag and skin friction losses in the inlet and nozzle where the Mach numbers tend to be quite high. A reasonable rule of thumb is that the stagnation pressure losses due to burner drag are comparable to the losses due to heat addition. 3.6 Ramjet control Let s examine what happens when we apply some control to the ramjet. The two main control mechanisms at our disposal are the fuel flow and the nozzle exit area. The engine we will use for illustration is a stable ramjet with an inlet shock and simple convergent nozzle shown below. The inlet throat is designed to have a Mach number well above one so that it is not so sensitive to un-start if the free stream conditions, burner temperature

17 CHAPTER 3. THE RAMJET CYCLE 3-7 or nozzle area change. Changes are assumed to take place slowly so that unsteady changes in the mass, momentum, and energy contained in the ramjet are negligible. The mass balance is Figure 3.4: Ramjet control model. ṁ e = + + 2( ) 2 P te A e p RTte f (M e )=(+f)ṁ a. (3.44) The pressure in the engine is virtually certain to be very high at this free stream Mach number, and so the nozzle is surely choked, and we can write ṁ a = ( + f) ( ) P te A e p RTte. (3.45) The thrust equation is T 2 = M 0 ( + f) U e + A e Pe. (3.46) P 0 A 0 U 0 A 0 P 0 Our main concern is to figure out what happens to the velocity ratio and pressure ratio as we control the fuel flow and nozzle exit area. Nozzle exit area control First, suppose A e is increased with T te constant. In order for (3.45) to be satisfied P te must drop keeping P te A e constant. The shock moves downstream to a higher shock Mach

18 CHAPTER 3. THE RAMJET CYCLE 3-8 number. The velocity ratio remains the same and, since the exit Mach number does not change, the product P e A e remains constant. Note that the thrust decreases. This can be seen by writing the second term in (3.46) as A e A 0 P e P 0 A e A 0. (3.47) The left term in (3.47) is constant but the right term increases leading to a decrease in thrust. If A e is decreased, the reverse happens, the inlet operates more e ciently and the thrust goes up. But remember, the amount by which the area can be decreased is limited by the Mach number of the inlet throat. Fuel flow control Now, suppose T te is decreased with A e constant. In order for (3.45) to be satisfied, P te must drop keeping P te / p T te constant. Once again the shock moves downstream to a higher shock Mach number. The velocity ratio goes down since the exit stagnation temperature is down and the Mach numbers do not change.the pressure ratio also decreases since the exit stagnation pressure is down. The thrust clearly decreases in this case. If T te is increased, the reverse happens, the inlet operates more e ciently and the thrust goes up. The amount by which the burner exit temperature can be increased is again limited by the Mach number of the inlet throat. 3.7 Example - Ramjet with un-started inlet For simplicity, assume constant heat capacity with =.4,C p = 005 M 2 / sec 2 K. The gas constant is R = 287 M 2 / sec 2 K. The ambient temperature and pressure are T 0 = 26K and P 0 =2 0 4 N/M 2. The fuel heating value is h f = J/kg. The sketch below shows a ramjet operating at a free stream Mach number of 3.0. A normal shock stands in front of the inlet. Heat is added between stations 3 and 4. Figure 3.5: Ramjet with normal shock ahead of the inlet.

19 CHAPTER 3. THE RAMJET CYCLE 3-9 The stagnation temperature at station 4 is T t4 = 2000 K. Relevant areas are A 3 /A.5 = 8, A = A 3 = A 4 and A 4 /A e = 3. Determine the dimensionless thrust T/(P 0 A ). Do not assume f. Neglect stagnation pressure losses due to wall friction and burner drag. Assume that the static pressure outside the nozzle has recovered to the ambient value. Suppose A.5 can be increased until A = A.5 = A 3. By what proportion would the air mass flow change? Solution The first point to recognize is that the stagnation pressure at station 4 exceeds the ambient by more than a factor of two - note the pressure outside the nozzle is assumed to have recovered to the ambient value. Thus the exit Mach number is one and the Mach number at station 4 is M 4 = The stagnation temperature at station 3 is T t3 = K. The fuel-air ratio is determined from the enthalpy balance For constant heat capacity ṁ f h f =(ṁ a +ṁ f ) h te ṁ a h t0. (3.48) f = T t4 T t0 h f C pt t0 = T t4 T t = (3.49) Now we need to determine the flow between stations and 3. To get started we will neglect the fuel addition for the moment. Knowing the Mach number at 4 and the stagnation temperatures at 3 and 4 we can use Rayleigh line results to estimate the Mach number at station 3. The stagnation temperature ratio across the burner is T t4 = T t4 T t T t3 T t M=0.975 T t4 T t3 = T t T t3 M=? T t3 M=? = = = = (3.50) The Rayleigh line tables give T t = T t3 M=? = ) M 3 =0.03. (3.5) This is a reasonable approximation to the Mach number at station 3. pressure ratio across the burner is The stagnation

20 CHAPTER 3. THE RAMJET CYCLE 3-20 P t4 = P t4 P t P t3 P t M=0.975 P t3 M=0.03 =.235 =0.98. (3.52).258 The subsonic critical Mach number for an area ratio of 8 is The fact that the Mach number at station 3 is higher than this value implies that there is a shock in the diverging part of the inlet and the inlet throat Mach number is equal to one. The stagnation pressure ratio between the inlet throat and the exit can be determined from a mass balance between stations.5 and e. P t.5 A.5 ( + f) p Tt.5 = P p tea e = Tte P te = r 2000 P t = (3.53) The results (3.52) and (3.53) determine the stagnation pressure ratio across the inlet shock and this determines the Mach number of the inlet shock. shock = =0.79 ) M shock =2.004 (3.54) Thus far the ramjet flow looks as shown in Figure 3.6. Figure 3.6: State I. The stagnation pressure ratio across the external shock is

21 CHAPTER 3. THE RAMJET CYCLE 3-2 and so the overall stagnation pressure ratio is P t = (3.55) P t0 M=3 The static pressure ratio is P te = P t P te = = (3.56) P t0 P t0 P t.5 The temperature ratio is The velocity ratio is P e = P te + 2 M 0 2 P 0 P t0 + 2 M e 2 T e = T te + 2 M! 0 2 T 0 T t0 +! = M e 2 = = (3.57) =7.72. (3.58).2 Across the inlet, the mass balance is U e = M r p e Te 7.72 = = (3.59) U 0 M 0 T 0 3 and so P t.5 A.5 = P t0 A 0 f (M 0 ) (3.60) Finally the thrust is A 0 A.5 = P t.5 = =.39. (3.6) P t0 f (M 0 ) T 2 = M A 0 A.5 0 ( + f) U e + A e Pe P 0 A A.5 A U 0 A P 0 (3.62) and

22 CHAPTER 3. THE RAMJET CYCLE 3-22 T P 0 A StateI = (/8) ( ) + (/3) (4.494 ) = State II = (3.63) Now increase the inlet throat area to the point where the inlet un-chokes. As the inlet throat area is increased the Mach number at station 3 will remain the same since it is determined by the choking at the nozzle exit and the fixed enthalpy rise across the burner. The mass balance between the inlet throat and the nozzle exit is again P t.5 A.5 ( + f) p Tt.5 = P p tea e. (3.64) Tte The stagnation pressure at station.5 is fixed by the loss across the external shock. The fuel-air ratio is fixed as are the temperatures in (3.64). As A.5 is increased, the equality (3.64) is maintained and the inlet shock moves to the left increasing P te. At the point where the the inlet throat un-chokes the shock is infinitely weak and the only stagnation pressure loss between station.5 and the nozzle exit is across the burner. This corresponds to A.5 StateII A e = +f A A.5 StateII = r P te State II Tt.5 = r =0.522 (3.65) P t.5 T te A A e A e A.5 StateII The mass flow through the engine has increased by the ratio ṁ astateii ṁ astatei = A.5 StateII A.5 StateI = A.5 StateII A A A.5 StateI At this condition the stagnation pressure ratio across the system is P te State II P t0 = Pt.5 Pte State II P t0 P t.5 = 3 = (3.66) = =.392 = A 0 StateII A 0 StateI. (3.67) = = (3.68)

23 CHAPTER 3. THE RAMJET CYCLE 3-23 The static pressure ratio is P estateii P 0 = P te State II + P t0 + 2 M M e 2! = = (3.69).2 The Mach number at station increases as A.5 increases and at the condition where the inlet is just about to un-choke reaches the same Mach number as station 3. At this condition the ramjet flow field looks like that shown in Figure 3.7. Figure 3.7: State II. The inlet shock is gone, the inlet Mach number has increased to M = M 3 and the external shock has moved somewhat closer to the inlet. Note that the capture area to throat area ratio is still A 0 StateII A.5 StateII = although both A.5 and A 0 have increased. Also P t.5 = =.39 (3.70) P t0 f (M 0 ) A 0 StateII A = A 0 StateII A.5 StateII =.39 = (3.7) A.5 StateII A The thrust formula is

24 CHAPTER 3. THE RAMJET CYCLE 3-24 T P 0 A StateII 2 A0 StateII = M 0 A.5 StateII A.5 StateII A ( + f) U e + A e PeStateII. U 0 A P 0 (3.72) The velocity ratio across the engine is unchanged by the increase in inlet throat area. The thrust of state II is T P 0 A StateII = ( ) (6.256 ) = =.624. (3.73) The reduced loss of stagnation pressure leads to almost a 60% increase in thrust at this condition. State III Now remove the inlet throat altogether. Now, suppose A.5 is increased until A = A.5 = A 3. The ramjet flow field looks like Figure 3.8. Figure 3.8: State III. With the inlet throat absent, the Mach number is constant between and 3. There is no change in mass flow, fuel-air ratio, stagnation pressure, or the position of the upstream shock. The capture area remains

25 CHAPTER 3. THE RAMJET CYCLE 3-25 A 0 StateIII A = A 0 StateII A = A 0 StateII A.5 StateII =.39 = (3.74) A.5 StateII A Therefore the thrust is the same as the thrust for State II, equation (3.74). If we want to position the upstream shock very near the entrance to the engine we have to increase the nozzle exit area and reduce the heat addition. State IV Open the nozzle exit fully. First increase the exit area to the point where A = A 3 = A 4 = A e. If we maintain T t4 = 2000 K the Mach number at station 4 becomes one and the Mach number between and 3 is, from the Rayleigh solution, M = M 3 = The ramjet flow at this condition is sketched in Figure 3.9. The velocity ratio is still Figure 3.9: State IV. U e = M r p e Te 7.72 = = (3.75) U 0 M 0 T 0 3 The stagnation pressure ratio across the burner is, from the Rayleigh solution, Across the whole system P te P t StateIV = (3.76)

26 CHAPTER 3. THE RAMJET CYCLE 3-26 P te P t0 StateIV = P t.5 P t0 StateIV P te P t.5 StateIV = = (3.77) and the static pressure ratio is P estateiv P 0 = P te State IV + P t0 + 2 M M e 2! =0.278 = (3.78).2 The area ratio is The thrust formula for state IV is A 0 StateIV = P tf (M ) = = (3.79) A P t0 f (M 0 ) T P 0 A StateIV 2 A0 StateIV = M 0 ( + f) U e + A e PeStateIV A U 0 A P 0 (3.80) which evaluates to T = ( ) + (5.274 ) = P 0 A StateIV = (3.8) Note the considerable increase in mass flow for state IV compared to state III. From (3.74) ṁ astateiv A0 StateIV = ṁ astateiii A A A 0 StateIII = =2.62 (3.82) which accounts for much of the increased thrust in spite of the increase in stagnation pressure loss across the burner. State V Reduce the burner outlet temperature until the shock is very close to station.

27 CHAPTER 3. THE RAMJET CYCLE 3-27 Now reduce T te until the Mach number at station 3 matches the Mach number behind the shock. From the Rayleigh solution, this occurs when T te = At this condition the ramjet flow field looks like Figure The static temperature ratio is Figure 3.20: State V. T e = T te + 2 M! 0 2 T 0 T t0 + 2 M e 2 The velocity ratio at this temperature is = =3.57. (3.83).2 U e = M r p e Te 3.57 = = (3.84) U 0 M 0 T 0 3 The stagnation pressure ratio across the burner at this condition is and across the system P te P t3 =0.889 (3.85) The static pressure ratio is P te P t0 = = (3.86)

28 CHAPTER 3. THE RAMJET CYCLE 3-28 P e = P te + 2 M 0 2 P 0 P t0 + 2 M e 2! = =5.67. (3.87).2 Neglecting the fuel flow, the thrust is (Note that at this condition A 0 = A = A e ). T 2 Ue Pe = M 0 + =.4 9 (0.629 )+(5.67 ) = = 0 P 0 A U 0 P 0 State VI Reduce the burner temperature slightly to establish supersonic flow up to the burner. (3.88) If the temperature at station 4 is decreased by an infinitesimal amount then supersonic flow will be established through the engine. Finally the flow is as shown in Figure 3.2. Figure 3.2: State VI. The mass flow, velocity ratio, pressure ratio and thrust all remain the same as in state V. If we were to reduce the fuel flow to the burner to zero we would be back to the state of undisturbed Mach three flow through a straight tube. 3.8 Very high speed flight - scramjets As the Mach number reaches values above 5 or so the ramjet cycle begins to become unusable and a new design has to be considered where the heat addition across the burner is carried out at supersonic Mach numbers. There are several reasons why this is so, all related to the very high stagnation temperature and stagnation pressure of high Mach number flight. To get started let s recall the thrust equation for the ramjet with a fully expanded nozzle.

29 CHAPTER 3. THE RAMJET CYCLE 3-29 T P 0 A = M 0 2 ( + f) M r! e Te M 0 T 0 (3.89) Define the thrust coe cient as C thrust = Thrust 2 0U 2 = 0 A 0 T 2 M 0 2 P 0 A 0. (3.90) If we assume ideal behavior (M e = M 0 ) the thrust coe cient becomes C thrust =2 ( + f) When we carried out the energy balance across the burner r! Tte. (3.9) T t0 ṁ f h f =(ṁ a +ṁ f ) h t4 ṁ a h t3 (3.92) we assumed that the fuel enthalpy was simply added to the flow without regard to the chemistry of the process. In fact the chemistry highly limits the range of fuel-air ratios that are possible. The stoichiometric reaction of JP-4 with air where the fuel and oxygen are completely consumed is CH O N 2! CO H 2 O N 2. (3.93) corresponding to a fuel-air ratio f = = (3.94) This is roughly the value of the fuel-air ratio that produces the maximum outlet temperature from the burner. If we choose the much more energetic hydrogen fuel, the reaction is corresponding to a fuel-air ratio H O N 2! H 2 O +.864N 2 (3.95)

30 CHAPTER 3. THE RAMJET CYCLE 3-30 f = The fuel enthalpies are generally taken to be = (3.96) h fjp h fh2 4 = J/kg = J/kg. (3.97) In the earlier discussion, we took the perspective that the ramjet cycle was limited by a red line temperature in the hot part of the engine. Let s relax this assumption and allow the maximum temperature to be free while keeping the fuel-air ratio constant. Furthermore let s continue to retain the assumption of constant heat capacities even at high Mach numbers. We will correct this eventually in Chapter 9, but for now we just want to see what happens to the ideal thrust coe cient (3.9) as we increase the free stream Mach number at constant fuel-air ratio. Using (3.92)), constant heat capacities, and assuming adiabatic flow in the inlet and nozzle we can express the stagnation temperature ratio across the engine as! T te f f = T t0 +f + 2 M f (3.98) where we recall that f = h f / (C p T 0 ). Now the thrust coe cient becomes 0 C thrust =2@ f ( + f) f + 2 M 0 2 +(+f)! /2 A. (3.99) The temperature ratio and thrust coe drag coe cient is defined as cient are plotted in Figures 3.22 and The C drag = Drag D 2 0U 2 = 0 A 0 2 M 0 2. (3.00) P 0 A 0 At high Mach numbers the drag coe cient of a body tends toward a constant value. For a sphere the drag coe cient tends toward a constant slightly less than one, about More streamlined bodies have lower drag and coe cients as low as 0.2 can be achieved. This observation together with Figure 3.23 indicates that as the Mach number increases it becomes harder and harder to produce thrust that exceeds drag. The thrust coe cient

31 CHAPTER 3. THE RAMJET CYCLE 3-3 Figure 3.22: Temperature ratio of an ideal ramjet at constant fuel-air ratio. Figure 3.23: Thrust coe cient of an ideal ramjet at constant fuel-air ratio.

32 CHAPTER 3. THE RAMJET CYCLE 3-32 drops below one at a Mach number of about 7 to 8 for both fuels. As the Mach number increases a conventional ramjet (even an ideal one) simply cannot produce enough thrust to overcome drag. The limiting thrust coe cient at infinite Mach number is lim C thrust = f. (3.0) M 0! This last result suggests that a scramjet can benefit from the choice of a fuel-rich mixture ratio as long as it does not exceed the flammability limit of the fuel. This would also provide additional fuel for cooling the vehicle. An advantage of a hydrogen system is that it can operate quite fuel rich. In addition, hydrogen has a higher heat capacity than any other fuel enabling it to be used to provide cooling for the vehicle that, in contrast to a re-entry body, has to operate in a very high temperature environment for long periods of time. The down side of hydrogen is that liquid hydrogen has to be stored at very low temperatures and the liquid density is only about /0 of that of JP-4. It is clear that small e ects can be important. For example when we developed the thrust formula we neglected the momentum of the injected fuel. In a realistic scramjet analysis that would have to be taken into account Real chemistry e ects The real chemistry of combustion shows that the problem is even worse than just discussed. The plot points in Figure 3.2 are derived from an equilibrium chemistry computation of the combustion of JP-4 with air. Notice that the temperature reached by the combustion gases is substantially less than the ideal, and for high Mach numbers the temperature rise from the reaction is actually less than one due to the cooling e ect of the added fuel. The solution to this problem, which has been pursued since the 960s, is to try to add heat with the burner operating with a supersonic Mach number on the order of two or so thereby cutting the static temperature of the air flowing into the combustor by almost a factor of two. This is the concept of a supersonically burning ramjet or scramjet. A generic sketch of such a system is shown in Figure Two Homework problems are used to illustrate basic concepts Scramjet operating envelope Figure 3.25 shows a widely circulated plot showing the altitude and Mach number regime where a scramjet might be expected to operate. Contours of constant free stream stagnation temperature and flow dynamic pressure are indicated on the plot.

33 CHAPTER 3. THE RAMJET CYCLE 3-33 Figure 3.24: Conceptual figure of a ramjet. Figure 3.25: Conceptual operating envelope of a scramjet.

34 CHAPTER 3. THE RAMJET CYCLE 3-34 This figure somewhat accurately illustrates the challenges of scramjet flight. These can be listed as follows. ) At high Mach numbers the vehicle is enveloped in an extremely high temperature gas for a long period of time perhaps more than an hour. 2)At high altittude, combustion is hard to sustain even at high Mach numbers because of the low atmospheric density and long chemical times. This defines the combustor blowout limit. 3) At lower altitude and high Mach number the free stream dynamic pressure increases to the point where the structural loads on the vehicle become untenable. Let s take a look at the vehicle structural limit (item 3) in a little more detail. This limit is presented as a line of constant dynamic pressure coinciding with increasing Mach number and altitude. But in supersonic flow, the free stream dynamic pressure is really not a su cient measure of the actual loads that are likely to act on the vehicle. Again let s make a constant heat capacity assumption and compare the free stream stagnation pressure to the free stream dynamic pressure as follows. Form the pressure coe cient P t P q = P 2 P M 2 + This function is plotted in Figure ! M 2 = + 2 M 2 2 M 2. (3.02) Figure 3.26: Comparison of stagnation pressure and dynamic pressure in supersonic flow. If one repeats this calculation with real gas e ects the stagnation pressures one calculates are even larger because of reduced values of. Consider the downstream end of the inlet where the flow enters the combuster. According to Figure 3.26 if, at a free stream Mach number of 8, the internal flow is brought isentropically

35 CHAPTER 3. THE RAMJET CYCLE 3-35 to low Mach number at the entrance to the combustor, the combuster will experience a pressure 28 times the free stream dynamic pressure. For example, at an altitude of about 26 kilometers and M 0 =8.0 a low Mach number combuster would operate at about 3200 psia (0,000 times the ambient atmospheric pressure at that altitude). This would require a very heavy structure making the whole idea impractical. If instead, the flow Mach number entering the combuster can be maintained at about Mach 2 then the combuster will operate at a much more feasible value of 28 bar or about 40 psia (somewhat higher with real gas e ects accounted for). This structural issue is at least as important as the thermochemistry issue in forcing the designer to consider operating the combuster at supersonic Mach numbers in order to attain hypersonic flight. 3.9 Problems Problem - Review -D gas dynamics with heat addition and area change. Consider the flow of a combustible gas mixture through a sudden expansion in a pipe shown in Figure Figure 3.27: Dump combustor. Combustion occurs between section at the exit of the small pipe and section 2 in the large pipe where the flow is uniform. Wall friction may be assumed to be negligible throughout. The flow at station has stagnation properties P t and T t. The Mach number at station, M is subsonic and the pressure on the annular step is approximately equal to P.The heat of reaction is denoted by Q. (i) Show that the exit Mach number is given by

36 CHAPTER 3. THE RAMJET CYCLE 3-36 M M M = (ii) Show that when Q = 0 and A 2 /A goes to infinity + Q 2 M + 2 M 2 C p T 2. (3.03) t A2 2 A + M P t = + P t2 2 M 2. (3.04) This limit is the case of a simple jet coming from an orifice in an infinite plane. Problem 2 - Show that for an ideal ramjet Do not assume f. th = 2 M M 0 2. (3.05) Problem 3 - Figure 3.28 shows a ramjet operating at a free stream Mach number of 0.7. Heat is added between stations 3 and 4 and the stagnation temperature at station 4 is T t4 = 000K. The Mach number at station 3 is very low. The ambient temperature and pressure are T 0 = 26 K and P 0 =2 0 4 N/M 2. Assume that f. Figure 3.28: Ramjet in subsonic flow. Using appropriate assumptions, estimate the dimensionless thrust T/(P 0 A 0 ) and the area ratio A 0 /A e. Problem 4 - Figure 3.29 shows a ramjet operating at a free stream Mach number M 0 =.5, with a normal shock in front of the engine. Heat is added between stations 3 and 4 and the stagnation temperature at station 4 is T t4 = 400 K. The Mach number at station 3

37 CHAPTER 3. THE RAMJET CYCLE 3-37 is very low. There is no shock in the inlet. The ambient temperature and pressure are T 0 = 26 K and P 0 =2 0 4 N/M 2. Assume that f. Figure 3.29: Ramjet in supersonic flow with a shock ahead of the inlet. Using appropriate assumptions, estimate the dimensionless thrust T/(P 0 A 0 ) and the area ratio A 0 /A e. Problem 5 - In Figure 3.30 a ramjet operates at a freestream Mach number of 3. The inlet is a straight duct and the Mach number of the flow entering the burner at station 3 is three. Heat is added across the burner such that the Mach number of the flow exiting the burner at station 4 is two. The ambient temperature and pressure are T 0 = 26 K and P 0 =2 0 4 N/M 2. Assume that f. Figure 3.30: Supersonically burning ramjet. The exit flow is expanded to P e = P 0. Determine the dimensionless thrust, T/(P 0 A 0 ). Problem 6 - In Figure 3.3 a ramjet operates at a freestream Mach number of 2.5. The ambient temperature and pressure are T 0 = 26 K and P 0 =2 0 4 N/M 2. The engine operates with a straight duct after the burner, A /A e =. Supersonic flow is established at the entrance of the inlet and a normal shock is stabilized somewhere in the diverging part of the inlet. The stagnation temperature exiting the burner is T t4 = 800 K. Neglect

38 CHAPTER 3. THE RAMJET CYCLE 3-38 wall friction. Determine the dimensionless thrust, T/(P 0 A 0 ). Figure 3.3: Ramjet with a constant area nozzle. Problem 7 - Figure 3.32 shows a ramjet test facility. A very large plenum contains Air at constant stagnation pressure and temperature, P t0, T t0. Jet fuel (h f = J/kg) is added between stations 3 and 4 ( A 3 = A 4 ) where combustion takes place. The flow exhausts to a large tank which is maintained at pressure P 0. Let P t0 /P 0 = 00. The upstream nozzle area ratio is A 3 /A.5 = 8. The exit area, A e can be varied in order to change the flow conditions in the engine. The gas temperature in the plenum is T t0 = K. To simplify the analysis, assume adiabatic flow, neglect wall friction and assume constant specific heat throughout with =.4. Figure 3.32: Ramjet test facility. Initially, the valve controlling A e is closed, A e = 0 and the fuel mass flow is shut o. Consider a test procedure where the nozzle area is opened, then closed. In the process A e /A.5 is slowly increased from zero causing Air to start flowing. The nozzle is opened until A e = A 3. Then the nozzle area is slowly reduced until A e /A.5 = 0 once again. Plot the thrust normalized by the plenum pressure and upstream throat area, T/(P t0 A.5 ) and the fuel-air ratio f as a function of A e /A.5. Distinguish points corresponding to increasing and decreasing A e. The fuel flow is adjusted to maintain the stagnation temperature at 4

39 CHAPTER 3. THE RAMJET CYCLE 3-39 at a constant value. Plot the results for three cases. T t4 = K(zero fuel flow) T t4 = 200 K (3.06) T t4 = 2700 K Neglect stagnation pressure loss across the burner due to aerodynamic drag of the burner, retain the loss due to heat addition. Do not assume f. What flight Mach number and altitude are being simulated at this condition? Problem 8 - In Figure 3.33 a ramjet operates at a free-stream Mach number of 3. The area ratio across the engine is A /A e = 2. Supersonic flow is established at the entrance of the inlet and a normal shock is stabilized in the diverging part of the inlet. The inlet throat Mach number is.0. The stagnation temperature exiting the burner is T t4 = 944 K. Assume =.4,R = 287 M 2 / sec 2 K, C p = 005 M 2 / sec 2 K. The ambient temperature and pressure are T 0 = 26 K and P 0 =2 0 4 N/M 2. Assume throughout that f. Figure 3.33: Ramjet with a simple convergent nozzle. i) Suppose the fuel flow is increased with the geometry of the engine held fixed. Estimate the value of f that would cause the inlet to unstart. Plot the dimensionless thrust, T/(P 0 A.5 ), specific impulse, and overall e ciency of the engine as a function of f. Plot your result beyond the point where the inlet un-starts and assume the engine does not flame out. Assume the fuel flow is throttled so as to keep the fuel/air ratio constant. Note that the thrust is normalized by a fixed geometric area rather than the capture area A 0 that changes when the engine un-starts. ii) With T t4 = 944 K, suppose A e is reduced keeping the fuel flow the same. Estimate the value of A /A e that would cause the inlet to un-start? Plot the dimensionless thrust, T/(P 0 A.5 ) specific impulse and overall e ciency of the engine as a function of A /A e. As in part (i), plot your result beyond the point where the inlet un-starts and assume the engine does not flame out.

40 CHAPTER 3. THE RAMJET CYCLE 3-40 Problem 9 - In Figure 3.34 a ramjet operates at a free-stream Mach number of 3. The area ratio across the engine is A /A e = 2. Supersonic flow is established at the entrance of the inlet and a normal shock is stabilized in the diverging part of the inlet. The stagnation temperature exiting the burner is T t4 = 944 K. The ambient temperature and pressure are T 0 = 26 K and P 0 =2 0 4 N/M 2. Do not assume that f. Do not neglect the e ects of wall friction. Figure 3.34: Ramjet with a convergent nozzle and inlet shock. i) Determine the fuel-air ratio, f. ii) Determine the dimensionless thrust, T/(P 0 A 0 ). Problem 0 - Assume =.4,R = 287 M 2 / sec 2 K, C p = 005 M 2 / sec 2 K. The ambient temperature and pressure are T 0 = 26 K and P 0 =2 0 4 N/M 2. Figure 3.35 shows a ramjet operating at a free stream Mach number of 3.0. The incoming air is decelerated to a Mach number of 2.0 at station 3. Figure 3.35: A scramjet concept. Heat is added between 3 and 4 bringing the Mach number at station 4 to one. The flow is then ideally expanded to P e = P 0. This type of engine with the combustion of fuel occurring in a supersonic stream is called a SCRAMJET (supersonic combustion ramjet). Determine the dimensionless thrust T/(P 0 A 0 ). Assume f if you wish and neglect stagnation pressure losses due to friction. Problem - A ramjet operates in the upper atmosphere at a high supersonic Mach number as shown in Figure Supersonic flow is established at the entrance of the

41 CHAPTER 3. THE RAMJET CYCLE 3-4 inlet and a normal shock is stabilized in the diverging part of the inlet. The stagnation temperature exiting the burner is su cient to produce substantial positive thrust. A small flat plate is placed downstream of the burner as shown, causing a small drop in stagnation pressure between station a and station b. The plate can be positioned so as to produce high drag as shown, or it can be rotated 90 so that the long dimension is aligned with the flow, producing lower drag. The engine operates with a convergent-divergent nozzle. The nozzle throat is choked and the nozzle exit is fully expanded. Figure 3.36: Ramjet with variable drag loss. Suppose the plate is rotated from the high drag position to the low drag position. State whether each of the following area-averaged quantities increases, decreases or remains the same. )M 3 5)T e 2)M a 6)P e 3)M b 7)U e 4)M e 8)Enginethrust Explain the answer to part 8) in terms of the drag force on the plate and the pressure forces that act on the engine inlet and nozzle. Problem 2 - In Figure 3.37 a ramjet operates at a freestream Mach number of 3. The area ratio across the engine is A /A e = 2. Supersonic flow is established at the entrance of the inlet and a normal shock is stabilized in the diverging part of the inlet. The stagnation temperature exiting the burner is T t4 = 52 K. The ambient temperature and pressure are T 0 = 26 K and P 0 =2 0 4 N/M 2. Do not neglect wall friction. i) Determine the dimensionless thrust, T/(P 0 A 0 ). ii) Suppose the wall friction is increased slightly. Determine if each of the following increases, decreases or remains the same.

42 CHAPTER 3. THE RAMJET CYCLE 3-42 Figure 3.37: Ramjet in supersonic flow. a) The Mach number at station 4 b) The Mach number at station.5 c) The shock Mach number d) The dimensionless thrust T/(P 0 A 0 ) Assume the inlet does not un-start. Problem 3 - In Figure 3.38 a ramjet operates at a freestream Mach number of 3. The area ratio across the engine is A /A 8 =.75. Supersonic flow is established at the entrance of the inlet and a normal shock is stabilized in the diverging part of the inlet. The stagnation temperature exiting the burner is T t4 = 52 K. The nozzle is fully expanded P e = P 0. The fuel enthalpy is h f = J/kg. The ambient temperature and pressure are T 0 = 26 K and P 0 =2 0 4 N/M 2. Figure 3.38: Ramjet with a converging-diverging nozzle. i) Determine the dimensionless thrust, T/(P 0 A 0 ). ii) Suppose T t4 is decreased slightly. Determine if each of the following increases, decreases or remains the same. a) The Mach number at station 4 b) The Mach number at station 3 c) The shock Mach number d) The nozzle exit pressure P e

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